Chapter 5: Problem 31
What is meant by an entbalpy change?
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Chapter 5: Problem 31
What is meant by an entbalpy change?
These are the key concepts you need to understand to accurately answer the question.
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Calculate \(\Delta E\) for a. the combustion of a gas that releases \(210.0 \mathrm{kJ}\) of heat to its surroundings and does \(65.5 \mathrm{kJ}\) of work on its surroundings. b. a chemical reaction that produces \(90.7 \mathrm{kJ}\) of heat but does no work on its surroundings.
Burning \(1.43 \mathrm{g}\) of methane \(\left[\mathrm{CH}_{4}(g), \text { molar mass } 16.0 \mathrm{g} /\right.\) mol] with excess \(\mathrm{O}_{2}\) raised the temperature of \(250 \mathrm{g}\) of water in a bomb calorimeter from \(22^{\circ} \mathrm{C}\) to \(98^{\circ} \mathrm{C}\). Calculate the molar enthalpy of combustion of methane.
The destruction of the ozone layer by chlorofluorocarbons (CFCs) can be described by the following reactions: $$\begin{aligned} \mathrm{ClO}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{Cl}(g)+2 \mathrm{O}_{2}(g) & & \Delta H_{\mathrm{rxn}}=-29.90 \mathrm{kJ} \\ 2 \mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g) & & \Delta H_{\mathrm{rxn}}=24.18 \mathrm{kJ} \end{aligned}$$ Determine the value of the heat of reaction for the following: $$\mathrm{Cl}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \quad \Delta H_{\mathrm{rxn}}=?$$
For which of the following reactions does \(\Delta H_{\mathrm{rxn}}^{\circ}\) also represent an enthalpy of formation? a. \(2 \mathrm{N}_{2}(g)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}_{2}(g)+2 \mathrm{NO}(g)\) b. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}(g)\) c. \(2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) d. \(\mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}_{2}(g)\)
Use the following data to determine whether the conversion of diamond into graphite is exothermic or endothermic: $$\begin{array}{rll} \mathrm{C}(s, \text { diamond })+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) & \Delta H^{\prime} & =-395.4 \mathrm{kJ} \\ 2 \mathrm{CO}_{2}(g) \rightarrow 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) & \Delta H^{\prime} & =566.0 \mathrm{kJ} \\ 2 \mathrm{CO}(g) \rightarrow \mathrm{C}(s, \text { graphite })+\mathrm{CO}_{2}(g) & \Delta H^{\circ} & =-172.5 \mathrm{kJ} \\ \mathrm{C}(s, \text { diamond }) \rightarrow \mathrm{C}(s, \text { graphite }) & \Delta H^{\circ} & =? \end{array}$$
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